Question

If a, b, c are three distinct real numbers in G.P. and a + b + c = xb, then prove that either x < -1 or x > 3.

Answer 1

Let r be the common ratio of G.P

a, b = ar, c = $a{r}^2$

Now, a + b + c = xb

$\Rightarrow a+ar+a{r}^2=x\left(ar\right)$

$\Rightarrow a(1+r+r^2)=xar$

$r^2+(1-x)r+1=0$

Here, r is real, so

$D\ge 0$

$(1-x{)}^2-4(1\left)\right(1)\ge 0$

$1+x^2-2x-4\ge 0$

$x^2-2x-3\ge 0$

$(x-3)(x+1)\ge 0$

$\Rightarrow x<-1$ or $x>3$

and $x\ne 3$ or - 1

[$\because \text{a, b and c are distinct real numbers}$]